A 70 kg student is riding on a skateboard at 4m/s. A 60 kg student is also riding a

t or f if an object has a canstant acceleration it must ve changing velocity at the same rate over over again

Angelina_Jolie [31]

The answer to your question is True

7 0

3 years ago

Four objects are situated along the y axis as follows: a 2.09-kg object is at +2.97 m, a 2.93-kg object is at +2.53 m, a 2.57-kg

Tcecarenko [31]

Answer:

x = 0 m

y = 1.02 m

Explanation:

M1 = 2.09 kg

y1 = 2.97 m

M2 = 2.93 kg

y2 = 2.53 m

M3 = 2.57 kg

y3 = 0 m

M4 = 3.92 kg

y5 = -0.496 m

since all objects are situated on the Y-axis, this means the x coordinate of the center of mass is 0.

To find the y coordinate of the center of mass, we apply the equation below.

sum of moment of the objects about the origin = moment of the total mass of objects about the center of mass

M1.y1 + M2.y2 + M3.y3 + M4.y4 = Mt.Y

(2.09 x 2.97) + (2.93 x 2.53) + (2.57 x 0) + (3.92 x -0.496) = (2.09 + 2.93 + 2.57 + 3.92) Y

11.68 = 11.51 Y

Y = 11.68 / 11.51 = 1.02 m

7 0

3 years ago

The distance between adjacent nodes in a standing wave pattern in a length of string is 25.0 cm:A. What is the wavelength of wav

mina [271]

A) 50 cm

B) 10000 cm/s

Explanation

Step 1

A)

If you know the distance between nodes and antinodes then use this equation:

$\begin{gathered} \frac{\lambda}{2}=D \\ \text{where}\lambda\text{ is the wavelength} \\ D\text{ is the distance betw}een\text{ nodes} \end{gathered}$

then, let

$D=\text{ 25 cm }$

now, replace to find the wavelength

$\begin{gathered} \frac{\lambda}{2}=25 \\ \text{Multiply both sides by 2} \\ \frac{\lambda}{2}\cdot2=25\cdot2 \\ \lambda=50\text{ Cm} \end{gathered}$

so, the wavelength is

A) 50 cm

Step 2

The speed of a wave can be found using the equation

$v=\lambda f$

or velocity = wavelength x frequency,

then,let

$\begin{gathered} \lambda=50\text{ cm} \\ f=200\text{ Hz} \end{gathered}$

replace and evaluate

$\begin{gathered} v=\lambda f \\ v=50\text{ cm }\cdot200\text{ HZ} \\ v=10000\text{ }\frac{\text{cm}}{s} \end{gathered}$

so

B) 10000 cm/s

I hope this helps you

6 0

1 year ago

After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 54.0 cm. The explorer finds that

Natasha2012 [34]

Answer:

g = 11.2 m/s²

Explanation:

First, we will calculate the time period of the pendulum:

$T = \frac{t}{n}$

where,

T = Time period = ?

t = time taken = 135 s

n = no. of swings in given time = 98

Therefore,

$T = \frac{135\ s}{98}$

T = 1.38 s

Now, we utilize the second formula for the time period of the simple pendulum, given as follows:

$T = 2\pi \sqrt{\frac{l}{g}}$

where,

l = length of pendulum = 54 cm = 0.54 m

g = acceleration due to gravity on the planet = ?

Therefore,

$(1.38\ s)^2 = 4\pi^2(\frac{0.54\ m}{g} )\\\\g = \frac{4\pi^2(0.54\ m)}{(1.38\ s)^2}$

3 0

2 years ago

This is a question on my physics test :)

Licemer1 [7]

Answer:

119.6 J/Kg°C

Explanation:

Data obtained from the question include:

Mass of substance (ms) = 170 g

Initial temperature of substance (Ts) = 120 °C

Volume of water = 200 mL

Initial temperature of water (Ts) = 10 °C

Temperature of the mixture (T2) = 12.6 °C

Density of water = 1 g/mL

Specific heat capacity of water (Cw) = 4200J/Kg°C

Specific heat capacity of substance (Cs) =..?

Next, we shall determine the mass of water. This can be obtained as follow:

Volume of water = 200 mL

Density of water = 1 g/mL

Mass of water =..?

Density = mass /volume

1 = mass /200

Cross multiply

Mass of water = 1 x 200

Mass of water = 200 g

Convert 200 g of water to Kg

Mass of water = 200/1000 0.2 Kg

Mass of water = 0.2 Kg

Now, we obtained the specific heat capacity of the substance using the following formula:

MwCw(T2 – Tw) + MsCs(T2 – Ts) = 0

Mass of water = 0.2 Kg

Initial temperature of water (Ts) = 10 °C

Specific heat capacity of water (Cw) = 4200J/Kg°C

Temperature of the mixture (T2) = 12.6 °C

Mass of substance (ms) = 170 g = 170/1000 = 0.17 Kg

Initial temperature of substance (Ts) = 120 °C

Specific heat capacity of substance (Cs) =..?

MwCw(T2 – Tw) + MsCs(T2 – Ts) = 0

0.2× 4200(12.6 – 10) + 0.17×Cs×(12.6 – 120) = 0

840(2.6) + 0.17Cs(– 107.4) = 0

2184 – 18.258Cs = 0

Rearrange

2184 = 18.258Cs

Divide both side by the coefficient of Cs i.e 18258

Cs = 2184/18.258

Cs = 119.6 J/Kg°C

Therefore, the specific heat capacity of the substance is 119.6 J/Kg°C

7 0

3 years ago